Problem: Let $a(x)=4x^4+2x^2-x+1$, and $b(x)=x^2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{4x^4+2x^2-x+1}{x^2}&=\dfrac{4 {x^4}+2 {x^2}}{ {x^2}}+\dfrac{-x+1}{x^2}\\\\ &={4x^2+2}+\dfrac{{-x+1}}{x^2}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${-x+1}$ is less than the degree of $x^2$, it follows that ${r(x)}={-x+1}$, and ${q(x)}={4x^2+2}$. To conclude, $q(x)=4x^2+2$ $r(x)=-x+1$ [Is there another way of doing this?]